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how to find ionisation potential
Actually ionisation potential means that "the amount of enrgy required to remove the electron from the valency shell of a particular atom".We have to find it as the depend upon its atomic size.It means that low size atoms have a low tendency to removing the electron.
The Ionization Potentials and Shielding of Electrons in the First ShellThe Bohr model of a hydrogen-like ion predicts that the total energy E of an electron is given byE = −Z²R/n²where Z is the net charge experienced by the electron, n is the principal quantum number and R is a constantequal to approximately 13.6 electron volts (eV). This formula is the result of the total energy beingequal to E = − Ze²/(2rn)where e is the charge of the electron and rn is the orbit radius when the principal quantumnumber is n. The orbit radius is given byrn = n²h/(Zmee²)where h is Planck''s constant divided by 2π and me is the mass of the electron.Shell StructureElectrons in atoms are organized in shells whose capacities are equal to 2m², where m is an integer. Thus therecan be at most 2 electrons in the first shell, 8 in the second shell and 8 in the third shell and 18 in each of the fourthand fifth shells. Here only the first shellis being considered. For elements above helium all of the electrons have been removed except two.Here are all the ionization potentials for such ions.The Ionization Potentials for the Electronsin the First Shell for the Elements Helium through CopperProtonNumberIonizationPotentialFirst ElectronIonizationPotentialSecond Electron2 54.41778 24.587413 122.45429 75.640184 217.71865 153.896615 340.2258 259.375216 489.99334 392.0877 667.046 552.07188 871.4101 739.299 1103.1176 953.911210 1362.1995 1195.828611 1648.702 1465.12112 1962.665 1761.80513 2304.141 2085.9814 2673.182 2437.6315 3069.842 2816.9116 3494.1892 3223.7817 3946.296 3658.52118 4426.2296 4120.885719 4934.046 4610.820 5469.864 5128.821 6033.712 5674.822 6625.82 624923 7246.12 6851.324 7894.81 7481.725 8571.94 8140.626 9277.69 882827 10012.12 9544.128 10775.4 10288.829 11567.617 11062.38The Ionization Potential of the First Electronas a Function of Proton NumberAn ion with only one electron is equivalent to the hydrogen atom but having a positivecharge of Z instead of one. The Bohr theory applies to such system. According to theBohr theory the ionization potential should beIE = Z²R/n²R is constant and n, the quantum number, is equal to 1. Thus the ionization potential should be proportionalto the proton number squared. Here is the plot of the relationship.It certainly looks like a quadratic relationship. To test this proposition more precisely the logarithms of theionization potential and the proton numbers are computed. The plot of these quantitites is shown below.The regression equation of the logartithm of ionization potential on the logarithm of the proton number isln(IE) = 2.604026184 + 2.004055992ln(#p)[2029.1] [4123.2] R² = 0.999998471The numbers in the square brackets are the t-ratios for the regression coefficients. For a regression coefficient to be statistically significantly different its magnitude must be greater than 2.0. As can be seen the regression coefficient forln(#p) is highly significant. And the value appears to be essentially 2, as the Bohr theory predicts. However this propositionis tested by comparing the difference between the regression coefficient and 2 with the standard deviation of the regressioncoefficient. Although the difference is small, 0.004056, the standard deviation of the regression coefficient is even smaller,0.000486039, the ratio is 8.3. Common sense however says that the Bohr theory is verified for the first electron in thethe first shell.The Z in the formula is the net charge experienced by the electron, the number of protons #p in the nucleus less anyshielding ε. The Bohr theory equation is thenIE = R(#p−ε)²which can be expressed asIE = R(#p² − 2#p*ε + ε²)where R is the Rydberg constant, 13.6 eV. Thus the appropriate regression equation would beIE = c0 + c1#p + c2(#p)²in whichc1<0c2 ≅ RSince in this case there seems to be no reason for there to be any shielding for the first electron the value of ε should be essentially zero.The results of the regression are: IE = 17.63611461 − 4.994593419#p + 13.89254281(#p)²[5.1] [-9.7] [858.9] R² = 0.999998221The coefficient of #p is negative but statistically different from zero. The coefficient of (#p)² is notably closeto 13.6. The value of ε can be found asε = ½(−c1/c2) = 0.179758072Somehow some small portion of the charge of the nucleus is being shielded. The Bohr theory works marvelously well for an atom or ion with one electron but not nearly so well for multiple electrons.One line of approach is to treat the atom or ion with two electrons as a many-body problem. This approach does not come upwith much in the way of definite results. An alternate line of approach is to say that electrons in the same shell shieldsome fraction of a positive charge in the nucleus. This shielding ratio would be based upon the proportion of the charge of an electron which is closer to the nucleus than the center of an electron. The value of one half would be plausible. However the shielding ratio could be affected by any deviation from spherical symmetry.The Shielding Ratios of the Electrons in the First Shellas a Function of Proton Number of the NucleusThe shielding ratio ρ for the electrons in the first shell can be computed asρ = 1 − E2(#p)/E1(#p)where E1 and E2 are the ionization potentials for the first and second electrons, respectively. The values for the elements for which the data is available in the CRC Handbook of Physics and Chemistry82nd Edition (2001-2002) are given below.The Shielding Ratio of the Electronsin the First Shellfor the Elements Helium through CopperProtonNumberShieldingRatio2 0.5481732263 0.3822986524 0.2931399775 0.2376380336 0.1998115737 0.1723632258 0.1516164439 0.13525883410 0.12213401911 0.11134880712 0.1023404413 0.0946821414 0.08811670915 0.08239251416 0.07738825417 0.07292281218 0.06898510219 0.06551337420 0.06235328721 0.05948444322 0.05687145123 0.05448709124 0.05232678225 0.05031999826 0.04847003927 0.04674534528 0.04515841629 0.043676844The graph of these data displays a very regular pattern.This appears to be a relationship of the formρ = α(#p)-βThe value of the parameters α and β can be found by plotting the logarithms of ρ and #p, as below.A regression of ln(ρ) on ln(#p) gives:ln(ρ) = 0.093012504 − 0.956463724ln(#p)[16.2] [-439.2] R² = 0.999865247The results indicate that the shielding ratio is predictable on the basis of the proton number, but the shielding of positive charge in the nucleus by the electrons in the first shellis primarily important for small nuclides. The Ionization Potential for the Second Electron as a Function of the Proton NumberPreviously a regression equation of the formIE = c0 + c1#p + c2(#p)²was applied to the data for the first electron and it was found that there was a shielding of 0.179758072 units of charge.When the data for the ionization potentials of the second electron is used the regression results areIE = 20.41371337 − 21.54911763#p + 13.85998758(#p)² [6.2] [-44.6] [912.9] R² = 0.999998302Again the coefficient of #p is negative and the coefficient of (#p)² is notably close to the Rydberg constant of 13.6.The value of the shielding that is found from the coefficients is 0.777385892. The difference between this value and thevalue for the first electron is 0.59762782, indicating that the second electron shields about 0.6 of a unit charge of thenucleus. This is notably close to the value of 0.5 suggested by the simplest formulation of the notion of shielding byelectrons in the same shell.
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